The sum of the squares of all the roots of the equation $$x^{2} + |x + 4| + |x − 4| − 35 = 0$$ is

Given equation is,

$$x^{2} + |x + 4| + |x − 4| − 35 = 0$$

i.) When $$x>4$$,

$$x^2+x+4+x-4-35=0$$

or, $$x^2+2x-35=0$$

or, $$x^2+7x-5x-35=0$$

or, $$\left(x+7\right)\left(x-5\right)=0$$

so, $$x=5,-7$$

But $$x=-7$$ is not acceptable as $$x>4$$

So, only acceptable value is $$x=5$$

ii.) When $$-4\le x\le4$$

$$x^2+x+4-x+4-35=0$$

or, $$x^2-27=0$$

or, $$x=\pm\ \sqrt{\ 27}$$

But $$\sqrt{\ 27}=5.196$$

So, the obtained values of $$x$$ does not lie in this range. Hence, the case is rejected.

iii.) When $$x<-4$$,

$$x^2-x-4+4-x-35=0$$

or, $$x^2-2x-35=0$$

or, $$x^2-7x+5x-35=0$$

or, $$\left(x-7\right)\left(x+5\right)=0$$

or, $$x=7,-5$$

But $$x=7$$ is not acceptable as $$x<-4$$

So, only acceptable value is $$x=-5$$

So, the roots of the equation are $$x=5,-5$$

Sum of square of roots =$$\left(5\right)^2+\left(-5\right)^2=25+25=50$$